0399. Evaluate Division

https://leetcode.com/problems/evaluate-division

Description

You are given an array of variable pairs equations and an array of real numbers values, where equations[i] = [Ai, Bi] and values[i] represent the equation Ai / Bi = values[i]. Each Ai or Bi is a string that represents a single variable.

You are also given some queries, where queries[j] = [Cj, Dj] represents the jth query where you must find the answer for Cj / Dj = ?.

Return the answers to all queries. If a single answer cannot be determined, return -1.0.

Note: The input is always valid. You may assume that evaluating the queries will not result in division by zero and that there is no contradiction.

Example 1:

**Input:** equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
**Output:** [6.00000,0.50000,-1.00000,1.00000,-1.00000]
**Explanation:** 
Given: *a / b = 2.0*, *b / c = 3.0*
queries are: *a / c = ?*, *b / a = ?*, *a / e = ?*, *a / a = ?*, *x / x = ?*
return: [6.0, 0.5, -1.0, 1.0, -1.0 ]

Example 2:

**Input:** equations = [["a","b"],["b","c"],["bc","cd"]], values = [1.5,2.5,5.0], queries = [["a","c"],["c","b"],["bc","cd"],["cd","bc"]]
**Output:** [3.75000,0.40000,5.00000,0.20000]

Example 3:

**Input:** equations = [["a","b"]], values = [0.5], queries = [["a","b"],["b","a"],["a","c"],["x","y"]]
**Output:** [0.50000,2.00000,-1.00000,-1.00000]

Constraints:

• 1 <= equations.length <= 20

• equations[i].length == 2

• 1 <= Ai.length, Bi.length <= 5

• values.length == equations.length

• 0.0 < values[i] <= 20.0

• 1 <= queries.length <= 20

• queries[i].length == 2

• 1 <= Cj.length, Dj.length <= 5

• Ai, Bi, Cj, Dj consist of lower case English letters and digits.

ac

class Solution {
    Map<String, Double> value;
    Map<String, Set<String>> graph;
    Set<String> visiting;

    public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {
        // edge cases
        // illegal arguments

        value = new HashMap<>();
        // build graph
        graph = new HashMap<>();
        for (int i = 0; i < equations.length; i++) {
            String[] e = equations[i];
            if (!graph.containsKey(e[0])) graph.put(e[0], new HashSet<String>());
            if (!graph.containsKey(e[1])) graph.put(e[1], new HashSet<String>());
            graph.get(e[0]).add(e[1]);
            graph.get(e[1]).add(e[0]);
            // store value
            value.put(e[0]+e[1], values[i]);
            value.put(e[1]+e[0], 1.0 / values[i]);
        }

        // dfs find path, get result
        visiting = new HashSet<String>();
        double[] res = new double[queries.length];
        for (int i = 0; i < queries.length; i++) {
            String[] q = queries[i];
            double tmp = dfs(q[0], q[1]);
            res[i] = tmp;
        }

        return res;
    }

    private double dfs(String from, String to) {
        // exit
        if (!graph.containsKey(from)) return -1.0;
        if (from.equals(to)) return 1.0;

        visiting.add(from);
        double res = -1.0;

        // look around
        for (String next : graph.get(from)) {
            if (visiting.contains(next)) continue;
            // if (next.equals(to)) {
            //     res = value.get(from+next);
            //     break;
            // } else {
                double tmp = dfs(next, to);
                if (tmp != -1.0) {
                    res = value.get(from+next) * tmp;
                }
            // }
        }

        visiting.remove(from);
        return res;
    }
}

/*
map store value
dfs find path
*/